The Eilenberg-Zilber theorem says that for singular homology there is a natural chain homotopy equivalence:
$$S_*(X)\otimes S_*(Y) \cong S_*(X\times Y)$$
The map in the reverse direction is the Alexander-Whitney map. Therefore we obtain a map
$$S_*(X)\rightarrow S_*(X\times X) \rightarrow S_*(X)\otimes S_*(X)$$
which makes $S_*(X)$ into a coalgebra.
My source (Selick's Introduction to Homotopy Theory) then states that this gives $H_*(X)$ the structure of a coalgebra. However, I think that the Kunneth formula goes the wrong way. The Kunneth formula says that there is a short exact sequence of abelian groups:
$$0\rightarrow H_*(C)\otimes H_*(D) \rightarrow H_*(C \otimes D) \rightarrow \operatorname{Tor}(H_*(C), H_*(D)) \rightarrow 0$$
(the astute will complain about a lack of coefficients. Add them in if that bothers you)
This is split, but not naturally, and when it is split it may not be split as modules over the coefficient ring. To make $H_*(X)$ into a coalgebra we need that splitting map. That requires $H_*(X)$ to be flat (in which case, I believe, it's an isomorphism).
That's quite a strong condition. In particular, it implies that cohomology is dual to homology.
Of course, if one works over a field then everything's fine, but then integral homology is so much more interesting than homology over a field.
In the situation for cohomology, only some of the directions are reversed, which means that the natural map is still from the tensor product of the cohomology groups to the cohomology of the product. Since the diagonal map now gets flipped, this is enough to define the ring structure on $H^*(X)$.
There are deeper reasons, though. Cohomology is a representable functor, and its representing object is a ring object (okay, graded ring object) in the homotopy category. That's the real reason why $H^*(X)$ is a ring (the Kunneth formula has nothing to do with defining this ring structure, by the way). It also means that cohomology operations (aka natural transformations) are, by the Yoneda lemma, much more accessible than the corresponding homology operations (I don't know of any detailed study of such).
Rings and algebras, being varieties of algebras (in the sense of universal or general algebra) are generally much easier to study than coalgebras. Whether this is more because we have a greater history and more experience, or whether they are inherently simpler is something I shall leave for another to answer. Certainly, I feel that I have a better idea of what a ring looks like than a coalgebra. One thing that makes life easier is that often spectral sequences are spectral sequences of rings, which makes them simpler to deal with - the more structure, the less room there is for things to get out of hand.
Added Later: One interesting thing about the coalgebra structure - when it exists - is that it is genuinely a coalgebra. There's no funny completions of the tensor product required. The comultiplication of a homology element is always a finite sum.
Two particularly good papers that are worth reading are the ones by Boardman, and Boardman, Johnson, and Wilson in the Handbook of Algebraic Topology. Although the focus is on operations of cohomology theories, the build-up is quite detailed and there's a lot about general properties of homology and cohomology theories there.
Added Even Later: One place where the coalgebra structure has been extremely successfully exploited is in the theory of cohomology cooperations. For a reasonable cohomology theory, the cooperations (which are homology groups of the representing spaces) are Hopf rings, which are algebra objects in the category of coalgebras.
Best Answer
With field coefficients, homology and cohomology are dual to each other. The cohomology of a space is an algebra, the homology of a space is a coalgebra. When the space is a group (or loop space or $H$-space...) both its homology and cohomology are Hopf algebras.
"The literature" is full of computations of the cohomology of the classifying space $BG$. In some cases by the way, Hopf algebras help, for example if you compute with coefficients in $\mathbb{Q}$. Then graded Hopf algebras over the rational field are classified, which gives you a strong indication on what the cohomology of $G$ looks like, and a spectral sequence argument tells you that $H^*(BG, \mathbb{Q})$ is a polynomial ring, in the end. Details in McCleary's book A user's guide to spectral sequences.